# Math Help - Please check my work

1. ## Please check my work

find the lim as x approaches 0 for sin(sin(sin(kx)))/x

I used l'Hopital's rule and got this

lim as x approaches 0 for cos(sin(sin(kx)cos(sin(kx))kcos(kx)

2. Originally Posted by Nichelle14
find the lim as x approaches 0 for sin(sin(sin(kx)))/x

I used l'Hopital's rule and got this

lim as x approaches 0 for cos(sin(sin(kx)cos(sin(kx))kcos(kx)

If you add brackets so that they are properly balanced your limit looks
OK, but to me this looks like it goes to k.

Also numerical checks indicates the k is the right answer.

RonL

3. Originally Posted by Nichelle14
find the lim as x approaches 0 for sin(sin(sin(kx)))/x

I used l'Hopital's rule and got this

lim as x approaches 0 for cos(sin(sin(kx)cos(sin(kx))kcos(kx)

The reason this didn't work is that you are missing a term in the above. Let $f(x)=\frac{sin [ sin ( sin[kx] ) ]}{x}$ then

$f'(x) = \frac{cos [ sin ( sin [kx] ) ] cos ( sin[kx] ) cos[kx] k}{x}$ - $\frac{sin [ sin ( sin[kx] ) ]}{x^2}$

This expression gives a mess if you try to take the limit as x goes to 0. (You wind up doing an infinite number of L'Hopital rules.)

My suggestion is to use a nested series of approximations for small values of x in the sine function:
$\lim_{x \to 0}sin(x) \to x - (1/6)x^3$
and keep terms to third order in x.

So the numerator of your limit becomes:
$\lim_{x \to 0}sin [ sin ( sin[kx] ) ] \to \lim_{x \to 0} sin [ sin ( kx - (1/6)k^3x^3 ) ]$ $\to \lim_{x \to 0}sin [ \Big(kx - (1/6)k^3x^3 \Big) - (1/6) \Big(kx - (1/6)k^3x^3 \Big)^3 ]$

$\to \lim_{x \to 0}sin [ kx - (1/6)k^3x^3 - (1/6)k^3x^3 ]$ (to third order in x)

$\to \lim_{x \to 0}sin [ kx - (2/6)k^3x^3 ]$

$\to \lim_{x \to 0}[ \Big( kx - (2/6)k^3x^3 \Big) - (1/6) \Big( kx - (2/6)k^3x^3 \Big)^3]$ $\to \lim_{x \to 0}[kx - (2/6)k^3x^3 - (1/6)k^3x^3]$ $\to \lim_{x \to 0}[kx - (1/2)k^3x^3]$

Thus
$\lim_{x \to 0} \frac{sin [ sin ( sin[kx] ) ]}{x} \to \lim_{x \to 0} \frac{kx - (1/2)k^3x^3}{x}$ $\to k$

-Dan

4. Originally Posted by topsquark
The reason this didn't work is that you are missing a term in the above. Let $f(x)=\frac{sin [ sin ( sin[kx] ) ]}{x}$ then

$f'(x) = \frac{cos [ sin ( sin [kx] ) ] cos ( sin[kx] ) cos[kx] k}{x}$ - $\frac{sin [ sin ( sin[kx] ) ]}{x^2}$

This expression gives a mess if you try to take the limit as x goes to 0. (You wind up doing an infinite number of L'Hopital rules.)
In L'Hopitals rule you differentiate the top and bottom separately, and
the limit of the indeterminate form is the limit of the ratio of the derivatives.

As far as I can see the OP has found the ratio of derivatives correctly.
If you put the missing brackets in you will see that what she has does
in fact go to k as x goes to zero, as expected.

RonL

5. Originally Posted by CaptainBlack
In L'Hopitals rule you differentiate the top and bottom separately, and
the limit of the indeterminate form is the limit of the ratio of the derivatives.

As far as I can see the OP has found the ratio of derivatives correctly.
If you put the missing brackets in you will see that what she has does
in fact go to k as x goes to zero, as expected.

RonL
Oh dear, I'm blushing now. (Ahem!) I knew that. Really I did!

-Dan

6. I orginally thought it would be k, but couldn't prove my work.

I still don't see where I am going wrong. Can you explain a little more? I appreciate your help.

7. Originally Posted by Nichelle14
I orginally thought it would be k, but couldn't prove my work.

I still don't see where I am going wrong. Can you explain a little more? I appreciate your help.
You have a product of three cosines of things which go to zero and k, but
that means all of the cosines go to 1, and so the whole product goes to k.

RonL

8. Thanks. I see where I left off my brackets.
You are pretty smart.